Rock Paper Scissor
Two people are playing rock-paper-scissors. They both have the same strategy:
- If win, play the same move again the next round.
- Otherwise, play the move that defeats the opponent's last move.
In the first round, the match-up is rock versus paper. If they continue playing round after round, will a round ever end in a draw?
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9 solutions
That's a beautiful observation! This also works for a generalized version of RPS like Rock-Paper-Lizard-Scissor-Spock
What if the first round was a draw ?
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Then each player would play the move that beat the other's previous move, meaning a draw one round results in a draw the next. Hence, infinite draws. In fact, you can see that if they play the same move and use the same strategy, they will always play the same move as each other. So, if you meet a clone of yourself, playing rock paper scissors will be pretty useless.
Summary
The winner sticks to his answer, the looser does what will defeat exactly that answer. The result: the looser now wins and the former winner loses. They will keep switching sides indefinitely, with no draw.
Details
Below winners are listed in green, losers in red
R o c k S c i s s o r s S c i s s o r s P a p e r P a p e r P a p e r R o c k R o c k
At the end the Rock versus Paper repeats, even if in reverse, so the following steps will be again the same, and none of them will be a draw.
Note
The strategy as described is incomplete. It provides no answer to a draw.
The strategy doesn't seem incomplete to me; if they have a tie, they'll just be stuck in an infinite cycle of ties
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The strategy said what to do if you win and what to do if you lose. It did not say what to do if you have a tie. In that sense it is incomplete. It could be that they each do something random and different if they had a tie, and get a different result after that.
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Perhaps the question has been modified. It now states what to do in a win and what to do otherwise. Otherwise could be lose or tie.
there could be a tie in the first round.
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The problems specifies that "In the first round, the match-up is rock versus paper." That is not a tie.
it is not possible to draw any round.cz, there is a funny matter. the looser will win the next round.as it is mentioned there technically.
so, if they play even number rounds the match will be draw. but no round is gonna be draw.
How do you know that the loser will always win the next round?
What if they played an odd number of rounds, will there ever be a draw?
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Otherwise, play the move that defeats the opponent's last move. -so it is mentioned there.
and ,if they play odd number of rounds ,it will not be a draw match.
Rock > Scissors
Paper > Rock
Scissors > Paper
- R vs P
- S vs P
- S vs R
- P vs R
- P vs S
- R vs S
- R vs P (same as 1., hence it loops)
We'll always have a move and move that beats it on the next round so No .
First round is won so the loser knows what would be next. And hence loser in previous game will win the next one. A draw is not possible.
How can I post my own questions ?
R vs.P; S vs.P; S vs.R; P vs.R; P vs.S; ...infinity and beyond
Let x n = ( p 1 ( n ) , p 2 ( n ) ) denote player 1 and 2 moves in round n . x n ∈ { R , P , S } × { R , P , S }
Game goes:
x 1 = (R, P) -> (S, P) -> (S, R) -> (P, R) = x 4
then due to summetry (rename player 1->2, 2->1), it continues:
-> (P, S) -> (R, S) = x 6
Then, x 7 = x 1 , so 6-cycle restarts, so there will never be a draw because none of x n , n=1,...,6, corresponds to a draw.
The first round is not a draw.
If some round is not a draw, then the next round is not a draw either (because the winner will play the same move, and the loser will play the move that beats the winner's move).
Therefore, by induction, there will never be a draw.